1. Stoichiometry Unit 8/9

2. Stoichiometry The process of determining how much product is made or how much reactant is needed during a chemical reaction.

3. We need a balanced equation It will tell us 2 things: –Which substances to begin with (reactants) and which substances we’ll end with (products) –The ratio of particles involved - can be individual particles or moles

4. Balanced Equation = Mole ratio Example: 2 Mg + 1 O 2  2 MgO Can be thought of as: For every 2 moles of Mg burned, 1 mole of O 2 is consumed and 2 moles of MgO are produced. This allows us to make predictions about how much reactants we’ll need & how much of a product we can make with what we have.

5. Making Predictions 3 Stages to Consider: –Before: what we have before the reaction takes place –Change: how much of each substance actually changes –After: how much of each substance is present after the reaction

6. Sample Problem Hydrogen sulfide gas, which smells like rotten eggs, burns in air to produce sulfur dioxide and water. How many moles of oxygen gas would be needed to completely burn 2.4 moles of hydrogen sulfide?

7. Sample Problem Step 1: Write & balance the equation. 2 H 2 S + 3 O 2  2 SO 2 + 2 H 2 O Before Change After

8. Sample Problem Step 2: Fill in the before line with GIVEN information. 2 H 2 S + 3 O 2  2 SO 2 + 2 H 2 O Before 2.4 mol xs 0 mol 0 mol Change____ mol After ***Assume reactants you don’t have amounts for are present with more than enough for the reaction.

9. Sample Problem Step 3: Use the ratio of coefficients to determine the amount of CHANGE made. 2 H 2 S + 3 O 2  2 SO 2 + 2 H 2 O Before 2.4 mol xs 0 mol 0 mol Change -2.4 mol -3.6mol +2.4 mol +2.4 mol After

10. Sample Problem Step 4: Complete the table for what remains AFTER the reaction is complete 2 H 2 S + 3 O 2  2 SO 2 + 2 H 2 O Before 2.4 mol xs 0 mol 0 mol Change -2.4 mol -3.6mol +2.4 mol +2.4 mol After 0 mol xs mol 2.4 mol 2.4 mol

11. Sample Problem In this case, the desired answer is in moles. However, you can also determine what the mass is by converting it using molar masses (like we did in chapter 5). EX: 3.6 mol O 2 x(32.0 g/1mol) = 115.2 g O 2 are consumed